$\sum_{n=1}^\infty \frac1{2^n} $
Somehow this gives $a = \frac12$, $r = \frac12$ to put into the formula $\sum_{n=1}^\infty ar^{n-1}$.
whats a general method (if any) for finding $a$ and $r$ in the geometric series formula?
i know that if you put a series in the form $\sum_{n=1}^\infty ar^{n-1} $ you can find if it is divergent (if $|r| \ge 1$) or convergent (if $|r| < 1$) with the sum being $\frac{a}{1-r}$. I am not quite sure however how to obtain $a$ and $r$ from a series and put it into the general form of the geometric series.
my question is HOW do i find the values $a$ and $r$
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$\begingroup$We know, $\displaystyle\sum_{k=1}^{n} ar^{k-1}= a\dfrac{1-r^{n}}{1-r}$.
Now, $\displaystyle\sum_{k=1}^\infty \dfrac1{2^k} =\lim_{n\to\infty}\displaystyle\sum_{k=1}^n \dfrac1{2^k} =\lim_{n\to\infty}\displaystyle\sum_{k=1}^n \left(\dfrac12\right)\cdot{\left(\dfrac1{2}\right)}^{k-1}$.
This is in the form of the above formula with $a=\dfrac12$ and $r=\dfrac12$. So the expression becomes,
$\lim_{n\to\infty} \dfrac12\dfrac{1-\left(\frac12\right)^{n}}{1-\frac12} =\lim_{n\to\infty} 1-\left(\frac12\right)^n=1.$
If we instead know, $\displaystyle\sum_{k=1}^{\infty} ar^{k-1}=\dfrac{a}{1-r}$ for $|r|<1$,
Then again,
$\displaystyle\sum_{k=1}^\infty \dfrac1{2^k}=\displaystyle\sum_{k=1}^\infty \left(\dfrac12\right)\left(\dfrac12\right)^{k-1}$
which is in the form of the preceding formula with $a=\frac12$ and $r=\frac12$ (and $|r|<1$).
$\endgroup$ 10 $\begingroup$The number $r$ is the ratio of two consecutive terms in $\sum_{n=1}^\infty ar^{n-1}$. That is, we have $$r=\frac{ar^n}{ar^{n-1}}.$$ And when you have $r$, you should be able to get $a$.
NOTE: If you have a series $\sum b_n$ and the ratio $\frac{b_n}{b_{n-1}}$ depends on $n$, then you know that your series is NOT geometric.
$\endgroup$ 3 $\begingroup$well this is the simple answer to the simple question.
it appears from further research that $a$ is simply the first evaluation of the series and $r$ is $\frac{a_2}{a_1}$. which makes $a_1 = \frac12, a_2 = \frac14,$ so $\frac{a_2}{a_1} = \frac12$, therefor $r = \frac12$.
nowwwww I see, I was dividing $a_1$ by $a_2$, not the other way around, to get $2$ earlier.
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